# Proof that a random variable has exponential distribution.

Supose that $$X_1$$ is a continuous and positive (real) random variable with exponential distribution, namely $$P(X_1>t)=e^{-\lambda t}\quad t>0$$ Now suppose that $$X_2$$ is another continuous and positive (real) random variable such that for every $$s>0$$ we have $$P(X_2>t\mid X_1=s)=P(X_1>t)=e^{-\lambda t}$$

Now, why can I conclude that $$X_2$$ has exponential distribution and moreover that $$X_1$$ and $$X_2$$ are independent?

The claim comes from the book "S. Ross - Stochastic Processes (second edition)" at page 64 when the author proves that the interrarival times in a Poisson process are i.i.d. random variables.

Since $$e^{-\lambda t}$$ does not depend on $$s$$, we have for all $$t$$ $$P(X_2 > t) = \int_0^\infty \Pr(X_2 > t \mid X_1 = s) \lambda e^{-\lambda s}\,ds = e^{-\lambda t}.$$ Thus $$X_2$$ has exponential distribution with parameter $$\lambda$$ and the identity $$\Pr(X_2 > t \mid X_1 = s) = \Pr(X_1 > t) = \Pr(X_2 > t), \qquad t,s > 0$$ shows that $$X_1$$ and $$X_2$$ are independent.
• Dubious: Are you actually able to prove that the identity implies the independence of $X_1$ and $X_2$? – Did Aug 26 '14 at 22:54
• @Did: I guess I am missing something, but cannot we say that $\Pr(X_2 > t \mid X_1) = \Pr(X_2 > t)$, hence $\Pr(X_1 > t_1, X_2 > t_2) = E[1_{X_1 > t_1} \Pr(X_2 > t_2)] = \Pr(X_1 > t_1)\Pr(X_2 > t_2)$? – Siméon Aug 27 '14 at 8:01